函数f(x)=x^-x的单调递减区间是

f(x)=x^a-x导数为f'(x)=a*x^(a-1)-1假设a>0,（1）a为偶数时,f'(x)=a*x^(a-1)-1在R上单调递增.令f'(x)=a*x^(a-1)-1=0 得x=(1/a)^(1/(a-1))则f(x)=x^a-x在（-无穷,(1/a)^(1/(a-1))]上递减,在（(1/a)^(1/(a-1)),+无穷)...